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A299173 a(n) is the maximum number of squared consecutive positive integers into which the integer n can be partitioned.

%I A299173 #28 Feb 06 2018 08:35:56
%S A299173 1,0,0,1,2,0,0,0,1,0,0,0,2,3,0,1,0,0,0,0,0,0,0,0,2,0,0,0,3,4,0,0,0,0,
%T A299173 0,1,0,0,0,0,2,0,0,0,0,0,0,0,1,3,0,0,0,4,5,0,0,0,0,0,2,0,0,1,0,0,0,0,
%U A299173 0,0,0,0,0,0,0,0,3,0,0,0,1,0,0,0,2,4,0,0,0,5,6,0,0,0,0,0,0,0,0,1
%N A299173 a(n) is the maximum number of squared consecutive positive integers into which the integer n can be partitioned.
%C A299173 a(k^2)>=1, the inequality being strict if k is in A097812.
%H A299173 Robert Israel, <a href="/A299173/b299173.txt">Table of n, a(n) for n = 1..10000</a>
%e A299173 25 = 5^2 = 3^2 + 4^2 and no such partition is longer, so a(25) = 2.
%e A299173 30 = 1^2 + 2^2 + 3^2 + 4^2 and no such partition is longer, so a(30) = 4.
%e A299173 2018 = 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2 + 13^2 + 14^2 + 15^2 + 16^2 + 17^2 + 18^2 and no such partition is longer, so a(2018) = 12. (This special example is due to _Seiichi Manyama_.) - _Jean-François Alcover_, Feb 05 2018
%p A299173 N:= 200: # to get a(1)..a(N)
%p A299173 A:= Vector(N):
%p A299173 S:= n -> n*(n+1)*(2*n+1)/6:
%p A299173 M:= floor(sqrt(N)):
%p A299173 for d from 1 to M do
%p A299173   for b from d to M do
%p A299173     s:= S(b) - S(b-d);
%p A299173     if s > N then break fi;
%p A299173     A[s]:= d
%p A299173 od od:
%p A299173 convert(A,list); # _Robert Israel_, Feb 04 2018
%t A299173 terms = 100; jmax = Ceiling[Sqrt[terms]]; kmax = Ceiling[(3*terms)^(1/3)]; Clear[a]; a[_] = 0; Do[r = Range[j, j + k - 1]; n = r . r; If[k > a[n], a[n] = k], {j, jmax}, {k, kmax}]; Array[a, terms]
%Y A299173 Cf. A034705, A097812, A130052, A111044, A234304, A234311, A296338, A298467.
%K A299173 nonn,look
%O A299173 1,5
%A A299173 _Jean-François Alcover_, Feb 04 2018